(This article was original published on in Oct 2009 on my Wikipedia User page. I’m re-posting on my blog to bring my writing together in one place.)

Goodstein sequences are very long sequences which eventually reach 0, but run for so long that Peano arithmetic cannot be used to prove that they reach 0.

## Introduction

Let $$G(n)$$ be the Goodstein sequence starting with $$n$$ and ending at 0.

$G(3) = (2 + 1, 3, 3, 2, 1, 0)$ $G(4) = (2^2, 2 \cdot 3^2 + 2 \cdot 3 + 2, \dots, 2\cdot 23^2, \dots, (24 \cdot 2^{24} - 1)^2, \dots, 3, 2, 1, 0)$

Let $$g_n$$ be the base of the hereditary notation of the last term of $$G(n)$$ (Alternatively, it is the length of the Goodstein sequence + 1). We shall call these the Goodstein numbers.

$$n$$ $$g_n$$
0 2
1 3
2 5
3 7
4 $$3 \cdot 2^{402653211} - 1 = (24 \cdot 2^{24}) 2^{24 \cdot 2^{24}} - 1$$

Now, in fact, if $$f(n) = (n+1) 2^{n+1} - 1$$, then $$g_4 = f^3(2)$$. I show that this function has a meaning.

## Growth of Goodstein Numbers

Let us define a family of functions:

• $f_0(B) = B + 1$
• $f_k(B) = f_{k-1}^{B+1}(B)$

Ah ha,

• $f_1(B) = 2B + 1$
• $f_2(B) = (B+1) \cdot 2^{B+1} - 1$
• $f_3(2) = f_2^3(2) = g_4$

In fact:

$$n$$ $$g_n$$ $$G(n)$$
0 2
1 $$f_0(2)$$ $$(2^0, \dots)$$
2 $$f_1(2)$$ $$(2^1, \dots)$$
3 $$f_1(3)$$ $$(2^1 + 1, 3^1, \dots)$$
4 $$f_3(2)$$ $$(2^2, 3^3 - 1, \dots)$$
5 $$f_4(3)$$ $$(2^2 + 1, 3^3, 4^4 - 1, \dots)$$
6 $$f_6(5)$$ $$(2^2 + 2, 3^3 + 2, 4^4 + 1, 5^5, 6^6 - 1, \dots)$$
7 $$f_8(7)$$ $$(2^2 + 2 + 1, 3^3 + 3, 4^4 + 3, \dots, 7^7, 8^8 - 1, \dots)$$

Notice the pattern? If $$B^k$$ appears in $$G(n)$$ then $$g_n = f_k(B)$$ (where $$B$$ is the base and $$k<B$$). Likewise, if $$B^B$$ appears, then $$g_n = f_{B+1}(B)$$.

In fact, let’s rename our functions (here $$\omega$$ is a label, not a variable — in fact, it is actually an ordinal number, or generalized index, who’s properties I hope to take advantage of):

• $f_{\omega^0}(B) = B + 1$
• $f_{\omega^k}(B) = f_{\omega^{k-1}}^{B+1}(B)$

And add a new one:

• $f_{\omega^\omega}(B) = f_{\omega^{B+1}}(B)$

Thus, if we have a value of the form $$\alpha$$ at base $$B$$ in $$G(n)$$, then $$g_n = f_\alpha(B)$$.

## Derivation of Growth Function

Note: It turns out that the function I define here is a variant of the fast-growing hierarchy much like the Hardy hierarchy.

Goodstein talks about the hereditary form of a number and the unique ordinal number associated with each hereditary form. For example:

• $$266 = 2^5 + 2^3 + 2 = 2^{2^2 + 1} + 2^{2 + 1} + 2$$ is in form $$\omega^{\omega^\omega + 1} + \omega^{\omega + 1} + \omega$$

Let us identify the hereditary form with that ordinal number.

If $$N$$ has hereditary form $$\alpha$$ with base $$B$$, then let $$f_\alpha(B)$$ be the base at which the the Goodstein sequence starting at $$N$$ in base $$B$$ will end.

For some values of $$\alpha$$:

• $f_0(B) = B$
• $f_1(B) = B + 1$
• $f_k(B) = f_1^k(B) = B + k$
• $f_\omega(B) = f_B(B+1) = f_1^{B+1}(B) = 2B + 1$
• $f_{\omega \cdot k}(B) = f_\omega^k(B)) = (B + 1) 2^k - 1$
• $f_{\omega^2}(B) = f_{\omega \cdot B + B}(B+1) = f_{\omega \cdot B}(f_B(B+1)) = f_{\omega \cdot B}(f_\omega(B)) = f_{\omega \cdot (B+1)}(B) = f_\omega^{B+1}(B) = (B + 1) 2^{B + 1} - 1$
• $f_{\omega^2 \cdot k}(B) = f_{\omega^2}^k(B)) > 2 \uparrow \uparrow k$
• $f_{\omega^3}(B) = f_{\omega^2 \cdot B + \omega \cdot B + B}(B+1) = f_{\omega^2 \cdot B}(f_{\omega^2 \cdot B + B}(B+1)) = f_{\omega^2 \cdot B}(f_{\omega^2}(B)) = f_{\omega^2 \cdot (B+1)}(B) = f_{\omega^2}^{B+1}(B) > 2 \uparrow \uparrow (B+1)$
• $f_{\omega^k}(B) = f_{\omega^{k-1} \cdot (B+1)}(B) = f_{\omega^{k-1}}^{B+1}(B) > 2 \uparrow^{k-1} (B+1)$
• $f_{\omega^\omega}(B) = f_{\omega^{B+1}}(B) > 2 \uparrow^B (B+1) = 2 \to B + 1 \to B$
• $f_{\omega^\omega \cdot k}(B) = f_{\omega^\omega}^k(B) > 2 \to B + 1 \to k \to 2$

Note: $$h(n) = 3 \uparrow^n 3 < 2 \uparrow^n (n-1) = f_{\omega^\omega}(n-1)$$. So Graham’s number $$\mathcal{G} = h^{64}(4) < f_{\omega^\omega}^{64}(4) = f_{\omega^\omega \cdot 64}(4)$$.

Now $$G(12) = (2^{2+1} + 2^2, \dots, g_4^{g_4 + 1}, g_4 (g_4 + 1)^{g_4 + 1}, \dots)$$ where we remember that $$g_4 = f_{\omega^\omega}(2) = 3 \cdot 2^{402653211} - 1 > 64$$. So, $$g_{12} = f_{\omega^\omega \cdot g_4}(g_4 + 1) > f_{\omega^\omega \cdot 64}(4) > \mathcal{G}$$

• $f_{\omega^{\omega + 1}}(B) = f_{\omega^\omega \cdot B + \omega^B \cdot B + \cdots + B}(B+1) = f_{\omega^\omega \cdot B}(f_{\omega^{B+1}}(B)) = f_{\omega^\omega \cdot B}(f_{\omega^\omega}(B)) = f_{\omega^\omega \cdot (B + 1)}(B) = f_{\omega^\omega}^{B+1}(B) > 2 \to B + 1 \to B + 1 \to 2$
• $f_{\omega^{\omega + 1} \cdot k}(B) = f_{\omega^{\omega + 1}}^k(B) > 2 \to B + 1 \to k \to 3$
• $f_{\omega^{\omega + 2}}(B) = f_{\omega^{\omega + 1} \cdot B + \omega^\omega \cdot B + \omega^B \cdot B + \cdots + B}(B+1) = f_{\omega^{\omega + 1} \cdot B}(f_{\omega^{\omega+1}}(B)) = f_{\omega^{\omega + 1} \cdot (B + 1)}(B) = f_{\omega^{\omega + 1}}^{B+1}(B) > 2 \to B + 1 \to B + 1 \to 3$
• $f_{\omega^{\omega + k}}(B) = f_{\omega^{\omega + (k-1)} \cdot (B + 1)}(B) = f_{\omega^{\omega + (k-1)}}^{B+1}(B) > 2 \to B + 1 \to B + 1 \to k + 1$
• $f_{\omega^{\omega \cdot 2}}(B) = f_{\omega^{\omega + (B+1)}}(B) > 2 \to B + 1 \to B + 1 \to B + 2$

• $f_{\omega^{\omega^\omega}}(B) = f_{\omega^{\omega^{B+1}}}(B) >> 2 \to \overbrace{B + 1 \to \cdots \to B + 1}^{B^2 + 2}$

Now let’s look back at the table:

$$n$$ $$g_n$$ $$G(n)$$
0 $$f_0(2) = 2$$ $$(0, \dots)$$
1 $$f_1(2) = f_1(g_0) = 3$$ $$(1, \dots)$$
2 $$f_\omega(2) = f_\omega(g_0) = 5$$ $$(2^1, \dots)$$
3 $$f_{\omega+1}(2) = f_\omega(f_1(2)) = f_\omega(g_1) = 7$$ $$(2^1 + 1, 3^1, \dots)$$
4 $$f_{\omega^\omega}(2) = f_{\omega^\omega}(g_0)$$ $$(2^2, \dots)$$
5 $$f_{\omega^\omega + 1}(2) = f_{\omega^\omega}(f_1(2)) = f_{\omega^\omega}(g_1)$$ $$(2^2 + 1, 3^3, \dots)$$
6 $$f_{\omega^\omega + \omega}(2) = f_{\omega^\omega}(f_\omega(2)) = f_{\omega^\omega}(g_2)$$ $$(2^2 + 2, 5^5, \dots)$$
7 $$f_{\omega^\omega + \omega + 1}(2) = f_{\omega^\omega}(f_\omega(f_1(2))) = f_{\omega^\omega}(g_3)$$ $$(2^2 + 2 + 1, 3^3 + 3, \dots, 7^7, \dots)$$
8 $$f_{\omega^{\omega + 1}}(2) = f_{\omega^{\omega + 1}}(g_0) = f_{\omega^\omega}^{g_0 + 1}(2) = f_{\omega^\omega}^{g_0}(g_4)$$ $$(2^{2 + 1}, \dots, 2 \cdot g_4^{g_4}, \dots)$$
9 $$f_{\omega^{\omega + 1} + 1}(2) = f_{\omega^{\omega + 1}}(f_1(2)) = f_{\omega^{\omega + 1}}(g_1) = f_{\omega^\omega}^{g_1 + 1}(f_1(2)) = f_{\omega^\omega}^{g_1}(g_5)$$ $$(2^{2 + 1} + 1, 3^{3 + 1}, \dots, 3 \cdot g_5^{g_5}, \dots)$$
10 $$f_{\omega^{\omega + 1} + \omega}(2) = f_{\omega^{\omega + 1}}(f_\omega(2)) = f_{\omega^{\omega + 1}}(g_2) = f_{\omega^\omega}^{g_2 + 1}(g_2) = f_{\omega^\omega}^{g_2}(g_6)$$ $$(2^{2 + 1} + 2, \dots, 5^{5 + 1}, \dots, 5 \cdot g_6^{g_6}, \dots)$$
11 $$f_{\omega^{\omega + 1} + \omega + 1}(2) = f_{\omega^{\omega + 1}}(f_\omega(f_1(2))) = f_{\omega^{\omega + 1}}(g_3)$$ $$(2^{2 + 1} + 2 + 1, 3^{3 + 1} + 3, \dots, 7^{7 + 1}, \dots, 7 \cdot g_7^{g_7}, \dots)$$
12 $$f_{\omega^{\omega + 1} + \omega^\omega}(2) = f_{\omega^{\omega + 1}}(f_{\omega^\omega}(2))) = f_{\omega^{\omega + 1}}(g_4)$$ $$(2^{2 + 1} + 2^2, \dots, g_4^{g_4 + 1}, \dots, g_4 \cdot g_8^{g_8}, \dots)$$
13 $$f_{\omega^{\omega + 1} + \omega^\omega + 1}(2) = f_{\omega^{\omega + 1}}(f_{\omega^\omega}(f_1(2)))) = f_{\omega^{\omega + 1}}(g_5)$$ $$(2^{2 + 1} + 2^2 + 1, 3^{3+1} + 3^3, \dots, g_5^{g_5 + 1}, \dots, g_5 \cdot g_9^{g_9}, \dots)$$
14 $$f_{\omega^{\omega + 1} + \omega^\omega + \omega}(2) = f_{\omega^{\omega + 1}}(f_{\omega^\omega}(f_\omega(2)))) = f_{\omega^{\omega + 1}}(g_6)$$ $$(2^{2 + 1} + 2^2 + 2, \dots, 5^{5+1} + 5^5, \dots, g_6^{g_6 + 1}, \dots, g_6 \cdot g_{10}^{g_{10}}, \dots)$$
15 $$f_{\omega^{\omega + 1} + \omega^\omega + \omega + 1}(2) = f_{\omega^{\omega + 1}}(f_{\omega^\omega}(f_\omega(f_1(2))))) = f_{\omega^{\omega + 1}}(g_7)$$ $$(2^{2 + 1} + 2^2 + 2 + 1, \dots, 7^{7+1} + 7^7, \dots, g_7^{g_7 + 1}, \dots, g_7 \cdot g_{11}^{g_{11}}, \dots)$$
16 $$f_{\omega^{\omega^\omega}}(2) = f_{\omega^{\omega^\omega}}(g_0) = ?$$ $$(2^{2^2}, \dots, 2 \cdot n^{2 \cdot n^2 + 2 \cdot n + 2}, \dots)$$

## Some Examples

• $g_{16} = f_{\omega^{\omega^\omega}}(g_0) = f_{\omega^{\omega^{g_0+1}}}(g_0) = f_{\omega^{\omega^{g_0} \cdot (g_0 + 1)}}(g_0) = f_{\omega^{\omega^{g_0} \cdot g_0 + g_0 + 1}}(g_0) = f_{\omega^{\omega^{g_0} \cdot g_0 + g_0}}^{g_0 + 1}(g_0) = f_{\omega^{\omega^{g_0} \cdot g_0 + g_0}}^{g_0}(f_{\omega^{\omega^2 \cdot 2 + 2}}(g_0)) = ?$
• $\begin{array}{rcl} g_6 &=& f_{\omega^\omega + \omega}(2) = f_{\omega^\omega}(f_\omega(2)) \\ &=& f_{\omega^\omega}(5) = f_{\omega^6}(5) = f_{\omega^5}^6(5) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^5(f_{\omega^2}(5))))) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^5(2^{B+1} - 1)))) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^4(2^{2^{B+1} - 1} - 1)))) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^3(2^{2^{2^{B+1} - 1} - 1} - 1)))) \\ \end{array}$